The given function is \( f(x) = 3x^2 – 5x + 2 \) and we need to prove that it is continuous at \( x = 1 \)
For a function \( f(x) \) to be continuous at a point \( x = c \), the following three conditions must be satisfied:
- \( f(c) \) is defined;
- \( \lim_{x \to c} f(x) \) exists;
- \( \lim_{x \to c} f(x) = f(c) \).
Solution:
For the given function:
1) \( f(1) = 3(1)^2 – 5(1) + 2 = 0 \). Therefore, \( f(1) \) is defined.
2)
\( \displaystyle \lim_{x \to 1} f(x) \\ = \lim_{x \to 1} (3x^2 – 5x + 2) \\ = 3(1)^2 – 5(1) + 2 = 0 \\ \)3) \( \lim_{x \to 1} f(x) = f(1) \).
Since all 3 conditions are satisfied, the given function \( f(x) = 3x^2 – 5x + 2 \) is continuous at \( x = 1 \).
Please let me know in the comments if you find any error in this solution.